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\title{\heiti\zihao{2} 习题2.4}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{设 $\left\{a_{n}\right\}$ 单调递增， $\left\{b_{n}\right\}$ 单调递减, 且 $\lim _{n \rightarrow \infty}\left(b_{n}-a_{n}\right)=0$ \\证明 $: \lim _{n \rightarrow \infty} a_{n}, \lim _{n \rightarrow \infty} b_{n}$ 都存在且相等.}
\textbf{证}\quad
显然$a_{n}$有上界$b_{1}$和下界$a_{1}$,$b_{n}$有下界$a_{1}$和上界$b_{1}$,从而二者单调有界,所以其极限存在,所以由极限的四则运算可知,$\lim_{n \rightarrow +\infty}a_{n}=\lim_{n \rightarrow +\infty}b_{n}$.

\section{证明下列数列收敛并求其极限值}
\subsection{设数列 $\left\{a_{n}\right\}$ 满足 $0<a_{1}<1, a_{n+1}=a_{n}\left(2-a_{n}\right)$}
\textbf{解}\quad
可由归纳法易知$a_{n}<1$.而$\frac{a_{n+1}}{a_{n}}=2-a_{n}>1$,所以$a_{n}$单增.所以$a_{n}$单调有界,所以其极限存在,设为$a$,从而有$a=a(2-a)$,解得$a=1$.


\subsection{数列 $\sqrt{2}, \sqrt{2+\sqrt{2}}, \sqrt{2+\sqrt{2+\sqrt{2}}}, \cdots$}
\textbf{解}\quad
其单调有上界$2$,所以其极限存在,设为$a$,则$\sqrt{2+a}=a$,解得$a=2$.

\section{证明:若 $a_{n}>0,$ 且 $\lim _{n \rightarrow \infty} \frac{a_{n}}{a_{n+1}}=l>1,$ 则 $\lim _{n \rightarrow \infty} a_{n}=0 .$}
\textbf{证}\quad
$$
\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\frac{a_{1}}{l^{n-1}}=0
$$


\section{设 $0<x_{0}<\frac{1}{3}, x_{n+1}=x_{n}\left(2-3 x_{n}\right), n=0,1,2 \cdots,$ 求极限 $\lim _{n \rightarrow \infty} x_{n}$}
\textbf{解}\quad
由归纳法易得$a_{n}<\frac{1}{3}$.又$\frac{a_{n+1}}{a_{n}}=2-3a_{n}>1$,所以$a_{n}$单增.所以由单调有界收敛定理,$a_{n}$极限存在,记为$a$,则$a=a(2-3a)$,解得$a=\frac{1}{3}$



\section{设 $0<c<1, a_{1}=\frac{c}{2}, a_{n+1}=\frac{c}{2}+\frac{a_{n}^{2}}{2},$ 证明 $:\left\{a_{n}\right\}$ 收敛,并求其极限.}
\textbf{解}\quad

(1)若$\frac{c}{2}<1-\sqrt{1-c}$,则$0<a_{1}<1-\sqrt{1-c}$.

归纳:若$n=k$时有$0<a_{k}<1-\sqrt{1-c}$,则$n=k+1$时,有
$$
0<a_{k+1}\frac{c}{2}+\frac{a_{n}^{2}}{2}<\frac{c}{2}+\frac{1+1-c-2\sqrt{1-c}}{2}=1-\sqrt{1-c}
$$
\par 故$a_{n}$有上界和下界.

$2(a_{n+1}-a_{n})=a_{n}^2-2a_{n}+c$,当$0<a_{n}<1-\sqrt{1-c}$时大于$0$,所以极限存在,设为$a$.则$a^{2}-2a+c=0,a=1-\sqrt{1-c}$.

同理,当$\frac{c}{2}>1-\sqrt{1-c}$时,同理可得$a_{n}$极限为$1-\sqrt{1-c}$.

$\frac{c}{2}=1-\sqrt{1-c}$时,$a_{n}=1-\sqrt{1-c}$.

综上,$a_{n}$极限存在,且为$1-\sqrt{1-c}$



\section{求下列极限:}
\subsection{设 $a>0, x_{1}>0, x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right), n=1,2, \cdots,$ 求 $\lim _{n \rightarrow \infty} x_{n}$}
\textbf{解}\quad
$x_{n+1}=\frac{1}{2}\left[  \sqrt{x_{n}}-\frac{\sqrt{a}}{\sqrt{x_{n}}}  \right]^{2}+\sqrt{a}\geqslant \sqrt{a}$.

$x_{n+1}-x_{n}=\frac{1}{2}\left(  \frac{a}{x_{n}} - x_{n}  \right)\leqslant 0$.

$\therefore x_{n}单减有界$,设其极限为$\alpha(\alpha>0)$,则$\alpha=\frac{1}{2}\left( \alpha+ \frac{a}{\alpha}\right)$,解得$\alpha = \sqrt{a}$.


\subsection{设 $a>0, x_{1}>0, x_{n+1}=\frac{1}{3}\left(2 x_{n}+\frac{a}{x_{n}^{2}}\right), n=1,2, \cdots,$ 求 $\lim _{n \rightarrow \infty} x_{n}$}
\textbf{解}\quad
由递推公式可知,数列 $\left\{a_{n}\right\}$ 满足 $a_{n+1} \geqslant \frac{1}{3} 3 \sqrt[3]{a_{n} a_{n} \frac{a}{a_{n}^{2}}}=\sqrt[3]{a}$.

$a_{n+1}-a_{n}=\frac{1}{3}\left(2 a_{n}+\frac{a}{a_{n}^{2}}\right)-a_{n} \leqslant \frac{a-a_{n}^{3}}{3 a_{n}^{2}} \leqslant 0,$ 即 $\left\{a_{n}\right\}$ 单调递减.
所以 $\left\{a_{n}\right\}$ 收敛, 设 $\lim _{n \rightarrow \infty} a_{n}=A,$ 对等式 $a_{n+1}=\frac{1}{3}\left(2 a_{n}+\frac{a}{a_{n}^{2}}\right)$ 取极限, 可得 $A=\frac{2 A}{3}+\frac{a}{3 A^{2}},$ 则有 $A=\pm \sqrt[3]{a},$ 因为 $a_{n} \geqslant \sqrt[3]{a},$ 所以 $\lim _{n \rightarrow \infty} a_{n}=A=\sqrt{a}$.

\section{设$0 \leqslant x_{n} \leqslant 1,\left(1-x_{n}\right) x_{n+1} \geqslant \frac{1}{4}, \forall n \geqslant 1,$ 证明数列 $\left\{x_{n}\right\}$收敛并求其极限.}
\textbf{解}\quad
因为 $0<a_{n}<1,$ 所以 $0<1-a_{n}<1 ;$ 由 $a_{n+1}\left(1-a_{n}\right) \geqslant \frac{1}{4}$ 可推知 $a_{n+1} \geqslant \frac{1}{4\left(1-a_{n}\right)},$ 又因为
$a_{n+1}-a_{n} \geqslant \frac{1}{4\left(1-a_{n}\right)}-a_{n}=\frac{\left(1-2 a_{n}\right)^{2}}{4\left(1-a_{n}\right)} \geqslant 0,$ 即 $\left\{a_{n}\right\}$ 单调递增.显然其有界.
所以 $\left\{a_{n}\right\}$ 收敛,设 $\lim a_{n}=A,$ 对 $a_{n+1}\left(1-a_{n}\right) \geqslant \frac{1}{4}$ 的两边取极限,得 $A(1-A) \geqslant \frac{1}{4},$ 化简
得 $4 A^{2}-4 A+1 \leqslant 0,(2 A-1)^{2} \leqslant 0,$ 故 $A=\frac{1}{2},$ 即 $\lim _{n \rightarrow \infty} x_{n}=\frac{1}{2}$

\section{求下列数列的极限:}
\subsection{$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n^{2}}\right)^{n^{2}}$}
\textbf{解}\quad
$\mathrm{e}$.

\subsection{$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n-2}\right)^{n}$}
\textbf{解}\quad
$\mathrm{e}$.

\subsection{$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}-\frac{1}{n^{2}}\right)^{n}$}
\textbf{解}\quad
$1+\frac{1}{n+2}<1+\frac{1}{n}-\frac{1}{n^{2}}<1+\frac{1}{n}$.由夹逼定理易得极限为$\mathrm{e}$.

\subsection{$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{\frac{1}{n}} $}
\textbf{解}\quad
$\lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty}\left(\frac{n-1}{n}\right)^{\frac{1}{n}}=\frac{\lim _{n \rightarrow \infty} \sqrt[n]{n-1}}{\lim _{n \rightarrow \infty} \sqrt[n]{n}}=1$

\section{求下列极限:}
\subsection{$\lim _{n \rightarrow \infty} \frac{\ln n}{n}$}
\textbf{解}\quad
$\mathrm{Stolz}$,$0$.

\subsection{$\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{\ln n}$}
\textbf{解}\quad
$$
\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n+1}}{\ln (n+1)-\ln n}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n+1}}{\ln \left(1+\frac{1}{n}\right)}
$$
\par 由极限的保序性
$$
\lim _{n \rightarrow \infty} \frac{\frac{1}{n+1}}{\frac{1}{n+1}} \leqslant \lim _{n \rightarrow \infty} \frac{\frac{1}{n+1}}{\ln \left(1+\frac{1}{n}\right)} \leqslant \lim _{n \rightarrow \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}}
$$
\par 所以 $\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{\ln n}=1 .$

\section{求极限$\lim _{n \rightarrow \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)$}
\textbf{解}\quad
$$
\lim _{n \rightarrow \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n}\right)=\int_{1}^{2}\ln x \mathrm{~d}x=\left.\ln x \right|_{1}^{2}=\ln 2
$$

\section{证明:序列 $x_{n}=1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}-2 \sqrt{n}$ 的极限存在.}
由于
$$
\frac{1}{2 \sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}
$$
\par 因为 $a_{n+1}-a_{n}=\frac{1}{\sqrt{n+1}}-2 \sqrt{n+1}+2 \sqrt{n}=\frac{1}{\sqrt{n+1}}-2(\sqrt{n+1}-\sqrt{n})<0,$ 所以数列单调递减.
$$
\begin{aligned}
a_{n}>1+2(\sqrt{n+1}-\sqrt{2}-\sqrt{n}) &=1-2 \sqrt{2}+2(\sqrt{n+1}-\sqrt{n}) \\
&=1-2 \sqrt{2}+2 \frac{1}{\sqrt{n+1}+\sqrt{n}}>-2 \sqrt{2}
\end{aligned}
$$
\par 即数列 $\left\{a_{n}\right\}$ 单调递减有下界,所以 $\left\{a_{n}\right\}$ 收敛.

\textbf{\textcolor{red}{注}} \quad 对于$k$次幂的数列的差分我们一般用$k-1$次幂的形式来放缩,例如本题的$\frac{1}{2}$次幂被用$-\frac{1}{2}$次幂的形式来放缩.可以由有理式求导得到此结论,此为常见套路.
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